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I have a set $$A=\left[0, \frac {\pi}2 \right] \cap \left \{ x \in \mathbb{R} : \sin {x} \in \mathbb{R} \setminus \mathbb{Q} \right \} $$ and I need to prove that $A$ is Borel set. I really tried to it, but I have no more ideas.

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    $\begingroup$ Welcome to mse! You mention that you've really tried to do it, but you should say more. What have you tried? Do you have any ideas of your own? Once we have a better idea of exactly where you're struggling, we can help you better ^_^ $\endgroup$ Commented Jun 19, 2022 at 18:14

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Since $\left[0,\dfrac{\pi}{2}\right]$ is a closed set, it generates the Borel Algebra and therefore, it is borel. The other set contains isolated points, so it is the countable (since you can label each interval with a rational number and $|\mathbb{Q}|=\aleph_0$) union of singletons, therefore it is also closed. Then, $\left[0, \frac {\pi}2 \right] \cap \left \{ x \in \mathbb{R} : \sin {x} \in \mathbb{R} \setminus \mathbb{Q} \right \}$ is a closed set (finite intersection of closed sets is closed). Therefore, $A$ is a borel set.

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